According to the Wikipedia's article: "The Game of Life, also known simply as Life, is a cellular automaton devised by the British mathematician John Horton Conway in 1970."

Given a board with m by n cells, each cell has an initial state live (1) or dead (0). Each cell interacts with its eight neighbors

1. Any live cell with fewer than two live neighbors dies, as if caused by under-population.
2. Any live cell with two or three live neighbors lives on to the next generation.
3. Any live cell with more than three live neighbors dies, as if by over-population..
4. Any dead cell with exactly three live neighbors becomes a live cell, as if by reproduction.

Write a function to compute the next state (after one update) of the board given its current state.

Follow up: 

1. Could you solve it in-place? Remember that the board needs to be updated at the same time: You cannot update some cells first and then use their updated values to update other cells.
2. In this question, we represent the board using a 2D array. In principle, the board is infinite, which would cause problems when the active area encroaches the border of the array. How would you address these problems?

Credits:

Special thanks to @jianchao.li.fighter for adding this problem and creating all test cases.

給定一個由0，1組成的矩陣，每一個元素表示一個細胞的存活，1存活，0死亡，其中下一次更新每個細胞的存活由上、下、左、右、左上、左下、右上、右下，八個細胞決定，存活規則如下：

     1、周圍存活細胞小於2個，則死亡。

    2、周圍存活細胞2個或3個，則存活到下一次更新。

    3、周圍存活細胞3個以上，則死亡。

    4、周圍存活細胞3個的死細胞，將會成為一個活細胞。

AC程式碼：

class Solution
{
public:
    void gameOfLife(vector<vector<int>>& board)
    {
        int rows=board.size();
        if(rows==0)
            return ;
        int colums=board[0].size();
        if(colums==0)
            return ;
        for(int i=0; i<rows; ++i)
        {
            for(int j=0; j<colums; ++j)
            {
                int sum=getNeighbor(board,rows,colums,i,j);
                if(sum==2)
                    continue;
                else if(sum==3)
                    board[i][j]=board[i][j]==0?3:1;
                else
                    board[i][j]=board[i][j]==1?2:0;
            }
        }
        for(int i=0;i<rows;++i)
        {
            for(int j=0;j<colums;++j)
                board[i][j]%=2;
        }


    }
private:
    int getNeighbor(vector<vector<int>>& board,int rows,int colums,int x,int y)
    {
        int sum=0;
        for(int i=x-1; i<x+2; ++i)
        {
            for(int j=y-1; j<y+2; ++j)
            {
                if(i==x&&j==y)
                    continue;
                if(i>=0&&i<rows&&j>=0&&j<colums&&(board[i][j]==1||board[i][j]==2))
                    ++sum;
            }
        }
        return sum;
    }
};